SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Thick-walled cylinder (Lamé)
Use this for “A cylinder with 100 mm inner and 125 mm outer radius holds 30 MPa inside. Find the hoop stress at the bore and compare the thin-wall value.”
Diagram
σavg
=
σₓ + σᵧ
2
=
137 + −30
2
σavg = 53.3 MPa
R
=
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
=
83.3
MPa.
σ₁
=
σavg
+
R
=
137
MPa;
σ₂
=
σavg
−
R
=
−30
MPa;
τₘₐₓ
=
R.
tan
2θₚ
=
2τₓᵧ
÷
(σₓ
−
σᵧ),
so
θₚ
=
0°:
the
turned
element
carries
σ₁
on
the
faces
turned
θₚ
and
no
shear.
τ
is
plotted
positive
down,
so
X
turns
to
σ₁
by
2θₚ
the
way
the
element
turns.
This is an example. Type your own values to start a new problem.
σθ (MPa)
=
p (MPa)
×
(
rₒ (mm)
)
2
+
(
rᵢ (mm)
)
2
(
rₒ (mm)
)
2
−
(
rᵢ (mm)
)
2
σₜₕᵢₙ (MPa)
=
p (MPa)
×
rᵢ (mm)
rₒ (mm)
−
rᵢ (mm)
rᵢ and rₒ are the cylinder’s inner and outer radii. The element sits at the bore: the hoop stress σθ across x, the radial stress −p across y.
σᵣ
Bore radial stress
worked out (not typed)
σₐ
Axial stress
calculated
σθ ÷ σₜₕᵢₙ
Lamé over thin-wall
worked out (not typed)
Formulas
σθ
=
p
×
rₒ
2
+
rᵢ
2
rₒ
2
−
rᵢ
2
136.6667
=
30
×
125
2
+
100
2
125
2
−
100
2
σₐ
=
p
×
rᵢ
2
rₒ
2
−
rᵢ
2
53.3333
=
30
×
100
2
125
2
−
100
2
σᵣ
=
−
p
−30 = −30
σₜₕᵢₙ
=
p
×
rᵢ
rₒ − rᵢ
120
=
30
×
100
125 − 100
σθ
σₜₕᵢₙ
=
σθ
σₜₕᵢₙ
1.1389
=
136.6667
120
Assumptions
•
Inside pressure only, the outside at 0 gauge; one elastic, uniform material.
•
The hoop stress is largest at the bore; the closed ends carry the pressure as σₐ.
•
σₜₕᵢₙ is pr ÷ t with the inside radius; it reads low as the wall thickens.
Step-by-step
WE KNOW
p = 30 MPa, rᵢ = 100 mm, rₒ = 125 mm
FIND
bore hoop stress (σθ), axial stress (σₐ), bore radial stress (σᵣ), thin-wall estimate (σₜₕᵢₙ), lamé over thin-wall (σθ ÷ σₜₕᵢₙ)
Step 1 · Find bore hoop stress (σθ)
Use σθ = p × (rₒ² + rᵢ²) ÷ (rₒ² − rᵢ²)
Lamé’s
hoop
stress
at
r
=
rᵢ,
with
σᵣ
=
−
p
at
the
bore
and
0
outside.
σθ
=
30
×
125
2
+
100
2
125
2
−
100
2
σθ
=
30
×
15,625 + 10,000
15,625 − 10,000
σθ
=
30
×
25,625
5,625
σθ
=
768,750
5,625
σθ = 136.6667 MPa
Step 2 · Find axial stress (σₐ)
Use σₐ = p × rᵢ² ÷ (rₒ² − rᵢ²)
The
end
cap’s
load
p
πrᵢ²
is
carried
by
the
wall’s
ring,
π(
rₒ
2
−
rᵢ
2
).
σₐ
=
30
×
100
2
125
2
−
100
2
σₐ
=
30
×
100
2
15,625 − 10,000
σₐ
=
30
×
100
2
5,625
σₐ
=
30
×
10,000
5,625
σₐ
=
300,000
5,625
σₐ = 53.3333 MPa
Step 3 · Find bore radial stress (σᵣ)
Use σᵣ = −p
At the bore the wall is pressed by the fluid: the radial stress is the pressure, in compression.
σᵣ = −30 MPa
Step 4 · Find thin-wall estimate (σₜₕᵢₙ)
Use σₜₕᵢₙ = p × rᵢ ÷ (rₒ − rᵢ)
The
thin-wall
rule
pr
t
with
t
=
rₒ
−
rᵢ,
to
compare
with
Lamé’s
value.
σₜₕᵢₙ
=
30
×
100
125 − 100
σₜₕᵢₙ
=
30
×
100
25
σₜₕᵢₙ
=
3,000
25
σₜₕᵢₙ = 120 MPa
Step 5 · Find lamé over thin-wall (σθ ÷ σₜₕᵢₙ)
Use σθ ÷ σₜₕᵢₙ = σθ ÷ σₜₕᵢₙ
Compare: how many times the thin-wall estimate Lamé’s bore stress is (above 1, the thin-wall rule reads low).
σθ
σₜₕᵢₙ
=
136.6667
120
σθ
σₜₕᵢₙ
=
1.1389
Check
Put the numbers back in. Both sides should match.
136.6667
=
30
×
125
2
+
100
2
125
2
−
100
2
✓
53.3333
=
30
×
100
2
125
2
−
100
2
✓
−30 = −30 ✓
120
=
30
×
100
125 − 100
✓
1.1389
=
136.6667
120
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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